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Question Number 28600 by A1B1C1D1 last updated on 27/Jan/18
Commented by abdo imad last updated on 27/Jan/18
from where come (3/8) sir ? there is no mistake in my response and if you find something inform me ...
$${from}\:{where}\:{come}\:\frac{\mathrm{3}}{\mathrm{8}}\:{sir}\:?\:{there}\:{is}\:{no}\:{mistake}\:{in}\:{my}\:{response} \\ $$$${and}\:{if}\:{you}\:{find}\:{something}\:{inform}\:{me}\:… \\ $$
Commented by abdo imad last updated on 27/Jan/18
= Σ_(k=1) ^∝ (k/((k^2 +2)^2 −4k^2 ))= Σ_(k=1) ^∝ (k/((k^2 +2 −2k)(k^2 +2+2k))) =(1/4)Σ_(k=1) ^∝ ( (1/(k^2 −2k+2)) −(1/(k^2 +2k+2))) let put S_n = Σ_(k=1) ^n ( (1/(k^2 −2k+2)) − (1/(k^2 +2k+2))) and v_k = (1/(k^2 −2k +2)) we have v_(k+1 ) = (1/((k+1)^2 −2(k+1) +2)) = (1/(k^2 +2k+1 −2k −2+2))= (1/(k^2 +1)) we have v_(k ) = (1/((k−1)^2 +1)) S_n = Σ_(k=1) ^n ( v_k −v_(k+1) ) = v_1 −v_2 +v_2 −v_3 +....v_n −v_(n+1) =v_1 −v_(n+1) = 1− (1/(n^2 +1)) lim S_n =1 so lim_(n→+∞) Σ_(k=1) ^n (1/(k^4 +4)) = (1/4)
$$=\:\sum_{{k}=\mathrm{1}} ^{\propto} \:\:\frac{{k}}{\left({k}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} \:−\mathrm{4}{k}^{\mathrm{2}} }=\:\sum_{{k}=\mathrm{1}} ^{\propto} \:\:\frac{{k}}{\left({k}^{\mathrm{2}} +\mathrm{2}\:−\mathrm{2}{k}\right)\left({k}^{\mathrm{2}} \:+\mathrm{2}+\mathrm{2}{k}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{k}=\mathrm{1}} ^{\propto} \left(\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \:−\mathrm{2}{k}+\mathrm{2}}\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}} \:+\mathrm{2}{k}+\mathrm{2}}\right) \\ $$$${let}\:{put}\:\:{S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{2}{k}+\mathrm{2}}\:−\:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{2}}\right)\:{and} \\ $$$${v}_{{k}} =\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{2}{k}\:+\mathrm{2}}\:{we}\:{have}\:{v}_{{k}+\mathrm{1}\:} =\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{2}\left({k}+\mathrm{1}\right)\:+\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \:+\mathrm{2}{k}+\mathrm{1}\:−\mathrm{2}{k}\:−\mathrm{2}+\mathrm{2}}=\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +\mathrm{1}}\:\:{we}\:{have}\:{v}_{{k}\:} =\:\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(\:{v}_{{k}} \:−{v}_{{k}+\mathrm{1}} \right)\:=\:{v}_{\mathrm{1}} \:−{v}_{\mathrm{2}} +{v}_{\mathrm{2}} −{v}_{\mathrm{3}} +….{v}_{{n}} \:−{v}_{{n}+\mathrm{1}} \\ $$$$={v}_{\mathrm{1}} −{v}_{{n}+\mathrm{1}} =\:\mathrm{1}−\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\:\:{lim}\:{S}_{{n}} =\mathrm{1}\:\:{so} \\ $$$${lim}_{{n}\rightarrow+\infty} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{4}} +\mathrm{4}}\:\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by A1B1C1D1 last updated on 27/Jan/18
The correct is (3/8) ′-′
$$\mathrm{The}\:\mathrm{correct}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{8}}\:'-' \\ $$
Commented by A1B1C1D1 last updated on 28/Jan/18
This was the feedback i was given... Thank you anyway.
$$\mathrm{This}\:\mathrm{was}\:\mathrm{the}\:\mathrm{feedback}\:\mathrm{i}\:\mathrm{was}\:\mathrm{given}… \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{anyway}. \\ $$

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